The point of intersection the lines (x-1)/2=(y-2)/3=(z-3)/4 and (x-4)…

$l_1: \frac{x-1}{2}=\frac{y-2}{3}=\frac{z-3}{4}$

$l_2 : \frac{x-4}{5}=\frac{y-1}{2} = z$

Let $l_2 : \frac{x-4}{5} = \frac{y-1}{2} = z = λ$

so  z = λ    y = 2λ + 1    x = 5λ+4

putting value in $l_1$

$\frac{5λ+4-1}{2} = \frac{2λ+1-2}{3}=\frac{λ-3}{4}$

⇒  $\frac{5λ+3}{2} = \frac{2λ-1}{3}=\frac{λ-3}{4}$

comparing any two we get

$\frac{2λ - 1}{3} = \frac{λ - 3}{4}$

⇒ 8λ - 4 = 3λ - 9

⇒ 5λ = -5

⇒  λ = -1

so x = 5(-1) + 4 = -1

y = 2(-1) + 1 = -1

z = -1

(-1, -1, -1)